Silver metal crystallises in a cubic closed packed arrangement with edge length 404 pm. Thus, radius of the silver atom is

$\mathrm{AB}=\mathrm{a},\mathrm{AC}=\mathrm{a}$

Radius = r, 

then $\mathrm{BC}=\mathrm{CD}+\mathrm{DE}+\mathrm{EB}=\mathrm{r}+2\mathrm{r}+\mathrm{r}=4\mathrm{r}$

For F.C.C or C.C.P  4r =$\sqrt{2}a$

$\mathrm{r}=\frac{\sqrt{2}}{4}\mathrm{a}$

$\therefore \mathrm{r}=\frac{\mathrm{a}}{2\sqrt{2}}=\frac{404}{2\sqrt{2}}=\frac{404}{2.828}=142.8\mathrm{pm}$

Note: Detailed calculations to relate radius with edge length have been given. Examiner can use direct relation.

 $$\begin{gathered} \mathrm{simple} \mathrm{cubic}, \mathrm{r}=\frac{\mathrm{a}}{2}; \\ \mathrm{hcp}/\mathrm{ccp},\mathrm{r}=\frac{\mathrm{a}}{2\sqrt{2}}; \\ \mathrm{bcc},\mathrm{r}=\frac{\sqrt{3}}{4}\mathrm{a} \end{gathered}$$